w**d
发帖数: 2334 | 1 What do you mean by 2^n? Is 'n' the order of quadrature ?
Could you be more specific?
For 1D, n classical (legendre) gauss points (& weights) will give
a quadrature with 2n order accuracy. If you don't care about the sign of
the weights, you can always find n-th order quadrature with any n distinct
points. To get the weights, you only need to solve a linear algebra equation.
So it is easy to make that 低阶时的积分点包含于高阶的积分点中.
But if you want all weights are non-negative, you might need to choose the |
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w**d
发帖数: 2334 | 2
If you don't care the sign of the weights. It should be easy. Just follow
the definition of the quadrature.
E.g., given n points, x1, ..., xn, assume the weights are w1, ..., wn,
If you want M-th order quadrature, i.e. for any fi = x^i, i=0, ..., M,
the following fact holds:(def of quadrature)
integral( fi ) = fi(x1)*w1 + .... + fi(xn) *wn
when x1, ..., xn are distinct and M = n-1. The weights always exist.
So if you don't care the sign of the weights, it is always possible
to choose 低阶的积分 |
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r****t
发帖数: 3 | 3 是的，n是the order of quadrature. 2^n是quadrature point的个数，当然不一定
非要是2^n,只要个数随n增长的不快就行。
我最关心的是低阶的积分点包含于高阶的积分点中。legendre积分点是自己随便取的吗？
如果我希望有比较多的积分点位于积分区间的两个端点处，能不能实现呢？
能不能推荐一本讲地比较好的书？谢谢
equation.
，
， |
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r****t
发帖数: 3 | 4 如果需要一种Gaussian quadrature scheme：低阶时的积分点包含于高阶的积分点中，
同时积分点的个数虽阶数的升高而增加得不快，比如以2^n的速度增加是可以接受的，
请问有没有这样的scheme? Thank you! |
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s**i
发帖数: 381 | 5 Is there any good reference on how to implement this?
I know Gaussian quadrature is much better for my particular case,I want to
use a Brute force trapzoidal rule.
Thanks. |
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w**d
发帖数: 2334 | 6 Is there such rule for tetrahedra?
You can check Ronald Cools' homepage for cubature.
He has the most complete reference on quadratures and cubatures, I think. |
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y**t
发帖数: 50 | 7 chebyshev quadrature is the guassian quardrature over the
integral [-1,1] with weighting functionW(x)=(1-x^2)^{-1/2}
The abscissas for quardrature n are given by the roots
of the chebyshev polynomial of the 1st kind T_n(x)
and there are formulas about the wights too
some example
n abscissas weights
2 +/-0.7071 1.5708
3 0 1.0472
+/-0.8660 1.0472
4...... |
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s********1
发帖数: 54 | 8 ______________________________________________________________________
In terms of the variance structure
______________________________________________________________________
Normal structure depend on each individual,Spatial structure depends on the
distances between two points and compound, etc.
______________________________________________________________________
In terms of the likelihood:
______________________________________________________________________
The PL method is based on Wol... 阅读全帖 |
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a**********u
发帖数: 28450 | 9 Charged-particle multiplicities in pp interactions at View the MathML source
measured with the ATLAS detector at the LHCstar, open, star, openstar, open
121a, 121b, S. Zenz14, D. Zerwas114, G. Zevi della Porta57, Z. Zhan32d, H.
Zhang83, J. Zhang5, Q. Zhang5, X. Zhang32d, L. Zhao107, T. Zhao137, Z.
Zhao32b, A. Zhemchugov65, S. Zheng32a, J. Zhong149, z, B. Zhou87, N. Zhou34,
Y. Zhou149, C.G. Zhu32d, H. Zhu41, Y. Zhu170, X. Zhuang98, V. Zhuravlov99,
B. Zilka143a, R. Zimmermann20, S. Zimmermann20, S... 阅读全帖 |
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o****l
发帖数: 21 | 10 Assume you have n interpolation points xi(i=1,2,3,....n)(n known f(xi)),
you divide the whole domain into many subdomain and each subdomain is defined
by two successive stationary points(the local maximum and minimum points).
Then you use Gaussian quadrature in each subdomain. Since all of this
can be done by computer, you do not need to worry about the tedious work.
As we know, m degree Gaussian quadrature gives us 2m-1 degree approximation.
So you can get your error estimation by defining it |
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m*******s
发帖数: 3142 | 11 现在遇到的问题要使用Kramers-Kronig relation的一个比较普遍的形式，如下图所示，
这个不是严格意义上的Hilbert transform，因为分母里面的z是复数，被积函数在整个
积分区间都
是非奇异的，不涉及Cauchy Principal Value。特别强调其中的g(x)是上半圆弧，只在
[-a,a]非
零，其余地方都是0，所以积分区间自然被限制到[-a,a]。
我暂时试验用Gaussian-Legendre quadrature来数值计算这个积分,
发现得到的结果的相对精度差不多只有10^(-4)，很难让人满意。即便是把格点数目加
倍，相对精度也
只能提高10^(-1)。没有想到用Gaussian-Legendre quadrature算这样一个普通的积分
竟然这么
差。
请问高手有没有比较好的方法来数值计算这种类型的积分？谢谢！ |
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h**********c
发帖数: 4120 | 12 三体问题是100年以前的问题，如前讨论，没有closed form solution,或者没有更好的
solution.陆续有人提出一些数列。把当年的论文又翻了翻。做数值解的，有个Q D
Wang，应该是老中；Stephen Smale,这个名气好像大些。1970s时候基本都是数值解，
数值解能解。
数值解就是先找 FIXED POINTS，从fixed point 算Floquet multiplier.好像是记得互
补的复根可可以算出一个周期性轨道，就是数值积分，guass quadrature,numerical
continuation.注意Floquet multiplier对七维方程这里怎么选我当时没弄明白。叔当
时根据能量守恒的观点，蒙到几个连接不同周期轨道，这些周期轨道是属于不同fixed
points.可以叫做 heteroclinic orbits.
基本是由Floquet multiplier来决定流形的变化，很多东西很恍惚，不是数学专业。周
期轨道，混沌，heteroclinic orbits在平衡点附近的特性。
解会有很多。有比较有趣的现象，比如invaria... 阅读全帖 |
|
s*******x
发帖数: 427 | 13 As the Embedded Software Engineer, you will be creating Freescale support
for VisSim. VisSim is a graphical language for simulation and model-based
embedded development. The core product, is used for general modeling,
simulation and control system design applications. With VisSim, you can
quickly develop virtual prototypes of any dynamic system.
Responsibilities:
• Create user interface to configure peripheral blocks.
• Port existing runtime support library to Freescale... 阅读全帖 |
|
c*******e
发帖数: 65 | 14 帮国内朋友发帖。谢谢~
Senior RF IC Designer ---- Nufront
Experienced RFIC designer contributed as a project leader of the RFIC
development group in Nufront, Beijing or Shanghai, China. The Chip design
focus is wireless radio transceiver for cellular systems. Experience in CMOS
RF& analog IC design is required. Knowledge of CMOS RF and analog IC debug
& characterization, device modeling, and design challenges in scaled CMOS
technologies as well as a strong academic and industry background is
strongly prefe... 阅读全帖 |
|
m****g
发帖数: 30 | 15 Senior RF IC Designer ---- Nufront
Experienced RFIC designer contributed as a project leader of the RFIC
development group in Nufront, Beijing or Shanghai, China. The Chip design
focus is wireless radio transceiver for cellular systems. Experience in CMOS
RF& analog IC design is required. Knowledge of CMOS RF and analog IC debug
& characterization, device modeling, and design challenges in scaled CMOS
technologies as well as a strong academic and industry background is
strongly preferred.
Respon... 阅读全帖 |
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y**********9
发帖数: 566 | 16 都是我自己的文章。本来是存在移动硬盘里面的，但是硬盘坏了，很多都恢复不了。所
以求大家帮忙下载。
我的电子邮件地址是 [email protected]
(function(){try{var s,a,i,j,r,c,l,b=document.getElementsByTagName("script");l=b[b.length-1].previousSibling;a=l.getAttribute('data-cfemail');if(a){s='';r=parseInt(a.substr(0,2),16);for(j=2;a.length-j;j+=2){c=parseInt(a.substr(j,2),16)^r;s+=String.fromCharCode(c);}s=document.createTextNode(s);l.parentNode.replaceChild(s,l);}}catch(e){}})();
/* ]]> */
。
非常感谢，有包子送
1. Yunchu Li, Lin Zhang, Muping Song, Bo Zhang, Jeng-Yu... 阅读全帖 |
|
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z****e
发帖数: 54598 | 18 Mathematics
While he is often regarded as a designer of mechanical devices, Archimedes
also made contributions to the field of mathematics. Plutarch wrote: "He
placed his whole affection and ambition in those purer speculations where
there can be no reference to the vulgar needs of life."[42]
Archimedes used the method of exhaustion to approximate the value of pi.
Archimedes was able to use infinitesimals in a way that is similar to modern
integral calculus. Through proof by contradiction (reduc... 阅读全帖 |
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v*******n
发帖数: 104 | 19 调了一晚加一天的有限元程序，终于发现程序本身是对的，是那个Q9单元有点
奇怪，用缩减积分2*2gauss quadrature 的时候我那道题竟然刚度阵奇异，而
用3*3就没有问题了。唉累死我了调程序就是这么没准儿。 |
|
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h**********c
发帖数: 4120 | 21 gauss quadrature, 不是太懂
有点得瑟了 |
|
c*******e
发帖数: 8624 | 22 gaussian quadrature
每个sub interval上点如果能多取一点的话
不知道能不能到需要的精度,tolerance可以这样取
两个不同精度,高的当作真值,点怎么取反正(-1,1)上
都可以查到,然后把每个sub interval map到(-1,1)
上面就可以了
不过恐怕不会很快 |
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c*****t
发帖数: 562 | 23 piece wise quadrature/cubic? |
|
c*******e
发帖数: 8624 | 24 kernal比如是f(x,y),x=x1(y) to x2(y), y=y1 to y2
我对matlab积分的函数不是很熟悉
首先是不是不能用dblquad,那样好象x也要求是定限?
我就想用int先积x部分,但是问题是int是符号积分
f(x,y)现在找不到这么一个解(等了很久也没结果)
平时我都自己写,4重的积分我也用Gaussian Quadrature
写过,但是效率好象很差,用的都是for loop,因为
这样的积分我要用很多次(在循环里面),是不是用
matlab本身的函数要效率好很多?
应该是很容易的问题,搞了1个多小时没搞出来,哪位帮忙看看吧. |
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w**d
发帖数: 2334 | 25 you can work out the exact integral.
But if you really want to use quadrature, check Ronald Cools' homepage. |
|
k****g
发帖数: 67 | 26 可是在每个单元里不是整个一块的polynomial，是piecewise的
所以用一般的quadrature方法都得不到精确解的。 |
|
c*******e
发帖数: 8624 | 27 gaussian quadrature
我以前做过4维的,都一样套就可以了 |
|
o****o
发帖数: 8077 | 28 要做一个很简单的数值积分问题，其实就是求如下一个简单幂函数的期望
F(U)=U^(a-1)， 实数 a > 0。
U的分布是一个Normal(\mu, \sigma) truncated at 0.
在对数据集中每个点进行这个计算的时候，出现了很奇怪的问题。我用的是quad，高斯
quadrature法
当abs(\mu) 远离0点， \sigma又较小的时候，即使\mu1 \mu2很接近计算出来的值也相差
很大。例如
>> quad(@numF1_e,1e-10,1-1e-10,0.00001,0, 3.0692,1.2014,0.5740,9.389589)
ans =
139.9154
>> quad(@numF1_e,1e-10,1-1e-10,0.00001,0, 3.0692,1.2014,0.5740,9.789589)
ans =
2.5851e-005
>> quad(@numF1_e,1e-10,1-1e-10,0.00001,0, 3.0692,1.2014,0.5740,9.689589)
ans =
149.5704
@numF1_e 是那个函数 |
|
s**i
发帖数: 381 | 29 Thanks. I checked his CUBPACK but it is only for 2D |
|
|
j*****u
发帖数: 82 | 31 Biological and Medical Imaging
做 FDTD Skin Model，
Three-Dimensional Imaging with Quadrature Microscope
Embryo Cell Counting
还有一些sensor detecting之类的
phd以后找工作好找么?
这类工作要身份么
谢谢 |
|
q*******n
发帖数: 52 | 32 请问有人熟悉这个吗？ 有什么好处？ 能简单地介绍一下吗？
不胜感激。我自己在网上搜了一下，找到的东西都很复杂。
还望GGJJ们给科普一下。感谢。 |
|
|
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m*********0
发帖数: 17 | 35 希望这里有大侠不吝指教。
正做一个1-bit Complex Quadradure Delta-Sigma ADC(Continuous-time 5-order),
用于无线. 已经完成了单channel的低通ADC设计, 可以达到~65dB SNR. 现准备加上
Coupling转换成带通Quadrature,碰到一些问题：
1）SNR degradation. 本以为低通到带通转换不会损失多少SNR, 可转换后降低了~30dB
. 不知道主要问题在哪儿。
2）看到几个Modulators输出饱和，但很难用Matlab去Rescale系数。好像Richard's
Toolbox只提供Discrete-time Rescaling, 对于Continuous-time的比较难办。特别这
个是Complex的双Channels.
3)没想通Excess Loop Delay. 感觉我的Loop Delay 已经是好多个时钟周期了，难道要
控制Excess Loop Delay到小于一个时钟周期(<10%?) 还有就是觉得Loop就像个Ring
Oscillator, 即使无信号输入也会... 阅读全帖 |
|
c*******e
发帖数: 65 | 36 帮国内朋友发帖。谢谢~
Senior RF IC Designer ---- Nufront
Experienced RFIC designer contributed as a project leader of the RFIC
development group in Nufront, Beijing or Shanghai, China. The Chip design
focus is wireless radio transceiver for cellular systems. Experience in CMOS
RF& analog IC design is required. Knowledge of CMOS RF and analog IC debug
& characterization, device modeling, and design challenges in scaled CMOS
technologies as well as a strong academic and industry background is
strongly prefe... 阅读全帖 |
|
|
l*****e
发帖数: 65 | 38
补充:
The book 'Special Functions' by Andrews, Askey and Roy.
page 253 All proofs are simple, easy to understand.
thm 5.4.1:Suppose that {Pn(x)} is a sequence of orthogonal polynomials w.r.t.
the distribution d\alpha(x) on the interval [a,b]. Then Pn(x) has n simple
zeros in [a,b].
remark: all roots are real. and no multiple roots. very very nice.
Thm 5.4.2 The zeros of Pn(x) and P_{n+1}(x) separate each other.
remark: in other words, the zeros interlace( 交错)
Thm 5.4.3 Let m |
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w**d
发帖数: 2334 | 39 It is a tough question if you want minimum number of points for given
accuracy. It is usually called cubiture(?).
A easiest way is to use the tensor product of one dimentional Gauss-Quadrature
points. |
|
z*b
发帖数: 12 | 40 yes, you need to approximate your curve with some smooth curve before yo
u do , for example quadratic functions (that's simpson's rule), but of c
ourse the accuracy might not be good enough, so you may use piecewise qu
adratic functions or in general consider spline functions. Another techn
ique called least squares fitting might also give a good smooth curve fo
r you to integrate using standard quadrature formulas.
numerical
looks |
|
G***p
发帖数: 59 | 41 why in physics text books, they always use summing in quadrature
to carry out the error analysis? what is the difference?
|
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f**n
发帖数: 401 | 42 The unknown function f(x,y) is defined in the area of x>=0 and y>=0, i.e.,
in the first quadrature. For any x,y, f(x,y) >= 0.
f(x,y) is characterized by the following PDE,
df(x,y)/dx + df(x,y)/dy = A(x)B(y) + C(x)D(y), where A and C are known
functions of x, and B and D are known functions of y.
This seems to be a inhomogenous first-order linear PDE. The function f looks quite similar to the wave equation, but the AB+CD term seems hard to handle. Is there a way to analytically solve the problem? |
|
w********s
发帖数: 59 | 43 I think GCQ (Gauss Chebshev quadrature) works well in your case. Do a
search. It is because the integrand tail vanishes very quickly, so the
partition of x should be chosen as non-uniform.
)/ |
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c*********n
发帖数: 128 | 44 There are a bunch of methods.
For example, google Gaussian quadrature integral method |
|
t*******y
发帖数: 11968 | 45 Could anyone provide a good approach to explain the symmetry in the weights
of quadrature formulas? Thanks. |
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h**********c
发帖数: 4120 | 46 quadrature,collocation, manifold
比较常用
interpolation, qed, etc etc. |
|
m*******s
发帖数: 3142 | 47 Dirac函数对数值结果没有意义，况且Dirac函数的导数本身需要在同别的函数积分的意
义下定义，反而增加运算量。
不如不进行Chebyshev多项式展开，换成其他比如Hermite多项式，Laguerre多项式等可
以进行fourier变换的多项式，是否可行？有没有成熟的算法？
上面那个求Fourier变换的想法来自一篇paper，里面说B(\omega)的Fourier变换应该是
一系列在时间域衰减的指数函数的和。而且B(\omega)的trace对\omega作图，通常是有
峰有谷的不规则曲线，所以paper上说f_1,f_2,f_3.....都是x的Lorentzian function
。我根本不知道该怎么实现这种拟合,通常的least square方法似乎用不上。所以我才
来这儿问。
不过你的想法也不错，不进行这种整体性的分解，而是element-wise看问题，约化到通
常的函数的Fourier变换。FFT算法似乎是理想的算法，不过就是我比较担心在\omega域
对B(\omega)采样可能有问题，因为B(\omega)的trace反映出B(\omega)强烈的不光滑性
... 阅读全帖 |
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s***e
发帖数: 267 | 48 If h is negative you can utilize normal cdf to bound?
If h is positive why taylor expansion does not work? It seems fine since x
is bounded. For numerical approximate solutions you can use quadrature
method or Monte Carlo. |
|
r****y
发帖数: 1437 | 49
For an integral
\int{a, b}{W(x)f(x)}
You can scale it to \int{-1, 1},
when W(x) = 1/sqrt(1-x^2), given N, we can calculate N points
in [-1, 1], get a set like this
x1, w1
x2, w2
...
xi, wi
...
And your integration can be approximated by
summation{i=1, N}{wi*f(xi)}
|
|
d*t
发帖数: 28 | 50 fun_1(theta) serves as coefficients together with the quadrature weights.
Then the problem can be computed by x=Fy where F is Fourier matrix
and y is the coefficiency vector, x is the result.
x=fft(y) |
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